Problem: Simplify the following expression: $y = \dfrac{9x^2- 41x- 20}{9x + 4}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(-20)} &=& -180 \\ {a} + {b} &=& &=& {-41} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-180$ and add them together. Remember, since $-180$ is negative, one of the factors must be negative. The factors that add up to ${-41}$ will be your ${a}$ and ${b}$ When ${a}$ is ${4}$ and ${b}$ is ${-45}$ $ \begin{eqnarray} {ab} &=& ({4})({-45}) &=& -180 \\ {a} + {b} &=& {4} + {-45} &=& -41 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 +{4}x) + ({-45}x {-20}) $ Factor out the common factors: $ x(9x + 4) - 5(9x + 4)$ Now factor out $(9x + 4)$ $ (9x + 4)(x - 5)$ The original expression can therefore be written: $ \dfrac{(9x + 4)(x - 5)}{9x + 4}$ We are dividing by $9x + 4$ , so $9x + 4 \neq 0$ Therefore, $x \neq -\frac{4}{9}$ This leaves us with $x - 5; x \neq -\frac{4}{9}$.